Inverse problem for semilinear wave equation with strong damping

Abstract

The initial-boundary and the inverse coefficient problems for the semilinear hyperbolic equation with strong damping are considered in this study. The conditions for the existence and uniqueness of solutions in Sobolev spaces to these problems have been established. The inverse problem involves determining the unknown time-dependent parameter in the right-hand side function of the equation using an additional integral type overdetermination condition.

1 Introduction

Propagation of sound in a viscous gas and other similar processes of the same nature can be described by the model hyperbolic equation of the third order, which includes a mixed derivative with respect to spatial and time variables

where η is a positive constant, and ηΔ_xu_t represents low viscosity.

Many important physical phenomena can be modeled with the use of Equation 1 and its generalizations. These are, in particular, processes that occur in viscous media (propagation of disturbances in viscoelastic and viscous-plastic rods, movement of a viscous compressible fluid, sound propagation in a viscous gas), wave processes in different media, acoustic waves in environments where wave propagation disrupts the state of thermodynamic and mechanical equilibrium, liquid filtration processes in porous media, heat transfer in a heterogeneous environment, moisture transfer in soils, and longitudinal vibrations in a homogenous bar with viscosity. The term Δ_xu_t indicates that the level of stress is proportional to the level of strains and to the strain rate [1–5].

Due to its wide range of applications, different problems for Equation 1 were investigated by many authors. For example, the unique solvability of the direct initial-boundary value problems for Equation 1 and its nonlinear generalizations with power nonlinearities have been studied in other research [1, 2, 4–11].

The inverse problems, with the integral overdetermination conditions, of identifying of the coefficients in the right-hand side function of hyperbolic equations without damping or for other types of equations have been investigated in many studies [12–18]. Their unique solvability has been solved with the use of the methods such as integral equations, the Green function, regularization, and the Shauder principle [14] and successive approximations [18]. The unique solvability of a two-dimensional inverse problem for the linear third-order hyperbolic equation with constant coefficients and with the unknown time-dependent lower coefficient has been proved in Mehraliyev et al. [19].

The main objective of this study is to determine the sufficient conditions for the existence and uniqueness of the solution to the inverse problem for the third-order semilinear hyperbolic equation with an unknown time- dependent function on its right-hand side. The unknown function is determined from the equation, subject to initial, boundary, and integral type overdetermination conditions. To prove the main results of the study, we use the properties of the solution for the corresponding initial-boundary value problem and the method of successive approximations. These results are new for semilinear n-dimensional third-order hyperbolic equations with non-constant coefficients and an unknown function on their right-hand side. The unique solvability of the initial-boundary value problem has been proved using of the method of Galerkin approximations and the methods of monotonicity and compactness.

2 Problem setting

Let Ω ⊂ ℝⁿ, n ∈ ℕ, be a bounded domain with the smooth boundary ∂Ω ∈ C¹ and 0 < T < ∞. Denote Q_τ = Ω × (0, τ), τ ∈ (0, T]; Q_{t1, t2} = Ω × (t₁, t₂), t₁, t₂ ∈ (0, T]. In this study, we consider the following inverse problem: find the sufficient conditions for the existence of a pair of functions (u(x, t), g(t)) that satisfies the equation with strong damping (in the sense of Definition 3.1).

and the initial, boundary, and overdetermination conditions

We shall use Lebesgue and Sobolev spaces L^∞(·), L²(·), H¹(·): = W^{1, 2}(·), C^k(·), C([0, T];L²(G)), (see, e.g., Gajewski et al. [20]).

Suppose that the data of the problem (2–5) satisfy the following conditions.

(H1): a_ij(x, t) = a_ji(x, t), b_ij(x, t) = b_ji(x, t), and

for all ξ ∈ ℝⁿ, almost all x ∈ Ω, all t ∈ [0, T], and i, j = 1, ..., n, where α₀, α₁ and β₀, β₁ are positive constants.

(H2): functions φ₁(x, ξ), φ₂(x, ξ) are measurable with respect to x ∈ Ω for all ξ ∈ ℝ¹ and continuously differentiable concerning ξ ∈ ℝ. Moreover,

for almost all x ∈ Ω and ξ, η ∈ ℝ, where L_i,0, L_i,1 are positive constants.

(H3): , ,

(H4): E ∈ C²([0, T]), , .

(H5): .

Denote

Let γ₀ = γ₀(Ω) be a coefficient in Friedrich's inequality.

3 Initial-boundary value problem

Definition 3.1. A function u(x, t) is considered to be a solution of problem 2–4 if , u satisfies (3), and

for all functions and τ ∈ (0, T].

Theorem 3.2. Under the assumptions (H1)–(H3) and g ∈ L²(0, T), a_ijt ≤ 0 for all i, j = 1, 2, …, n, the problem (2–4) has a unique solution.

Proof. First, using Galerkin method, we prove the existence of a solution for the problem. Let be a basis in , orthonormal in L²(Ω). We will consider the sequence of functions

where the set is a solution of the initial value problem

Here and

The solution of system (8) exists on some interval [0, τ₀] (Carathéodory's Theorem [21, p. 43]). The estimation (13) from below implies that this solution could be extended on [0, T]. Multiplying each equation of (8) on function respectively, summing up for k from 1 to N and integrating to t on interval [0, τ], τ ≤ τ₀, we obtain

After transformations of terms from (9), we get

Note that

then from (10) we obtain

We rewrite the last inequality in the form

where

Then by Grönwall's lemma, from (12), we get

Therefore, from (11) we also get

Multiplying each equation of (8) on function respectively, summing up with respect to k from 1 to N and integrating on interval [0, τ], τ ≤ T, we obtain

After transformations in all terms from (15), we get

where Taking into account (13), (14), from (16) we obtain

where

The right-hand sides of the estimates (13), (14), and (17) are positive constants, independent of N. Therefore, there exists a subsequence of (which will be denoted by the same notation), such that as N → ∞

It follows from (18) that and therefore, as N → ∞. Besides, , and weakly in

Equations 8 and 18 imply the equality

for all functions and τ ∈ (0, T].

Let us prove that χ = φ₂(x, u_t).

Note that for all k ∈ ℕ. Due to (18), is fundamental in . So, for any ε > 0, there exists such a number N₀ that for all N, k ∈ ℕ, N > N₀ the inequality holds; thus, is also fundamental in and, therefore,

Consider the sequence

where . From (9), it follows that

After substitution (22) in (21), passing to the limit as N → ∞, taking into account (18), (20), and the assumptions of Theorem 3.2, we obtain

Choosing here , dividing the result on ϰ and then tending ϰ → 0 we obtain χ = φ₂(x, u_t). Hence, from (19), it follows (7).

Now we prove the uniqueness of the solution for the problems (2–4). On the contrary, suppose that there exist two solutions u₍₁₎(x, t) and u₍₂₎(x, t) of problems (2–4). Then ũ: = ũ(x, t) = u₍₁₎(x, t)−u₍₂₎(x, t) satisfies the conditions ũ(x, 0) ≡ 0, ũ_t(x, 0) ≡ 0, and the equality

holds for all τ ∈ (0, T].

After choosing v = ũ_t in (23) we get

From (24) by the same way as from (11) we got (12), we find the following estimate

Then from Grönwall's lemma and (25) we obtain and hence, ũ ≡ 0, and, therefore, u₍₁₎ = u₍₂₎ in Q_T.

4 Inverse problem

Definition 4.1. A pair of functions (u(x, t), g(t)) is a solution to the problem (2–5), if , and g ∈ C([0, T]), and it satisfies (5) and

holds for all functions and τ ∈ (0, T].

4.1 The equivalent problem

In this section, we shall find the equivalent problem for the problem (2–5).

Lemma 4.2. Letthe assumptions of Theorem 3.2, (H4), and (H5) hold. A pair of functions (u(x, t), g(t)), where, g ∈ C([0, T]), is a solution to the problem (2–5) if and only if it satisfies (26) for all functionsand for τ ∈ (0, T], the equality

holds fort ∈ [0, T].

Proof. Necessity: Let (u(x, t), g(t)) be a solution to the problem (2–5). From (26) and (5), it follows that

By integrating by parts in (28) and using the condition (H4), we get the equality

From (29), we can obtain (27).

Sufficiency: Let g^* ∈ C([0, T]), , and they satisfy (4), (26), and (27). Then u^* is a solution to the problem (2–4) with g^* instead of g in (2).

We set t ≥ 0. In exactly the same way as in the proof of necessity, we obtain

On the other hand g^*(t) and u^*(x, t) satisfy (27)

It follows from (30), (31) that

Integrating (32) with the use of the equalities , implies E^*(t) = E(t), t ≥ 0. Hence, u^*(x, t) satisfies the overdetermination condition (5).

4.2 Main results

Let Denote

Theorem 4.3. Let a_ijt ≤ 0 for all i, j = 1, 2, …, n, and the assumptions (H1) – (H5) hold. Then there exists a unique solution to the problem (2–5).

Proof. I. In the first step, we shall prove the theorem for T ≤ T₀.

We construct an approximation (u^m(x, t), g^m(t)) of the solution of problem (2–5), where g¹(t): = 0, the functions g^m(t), m ≥ 2, satisfy the equality

and u^m satisfies the equality

for all , and the conditions

It follows from Theorem 3.2 that for each m ∈ ℕ there exists a unique function , that satisfies (34), (35). Now we show that converges to the solution of the problem (2–5). Denote

Equation 33 for t ∈ (0, T₀] and m ≥ 3, implies the equality

We square both sides of equality (36) and integrate the result with respect to t, taking into account the hypotheses (H5), then we obtain

Then (37) implies the estimate

It follows from (35) that Hence, from (34) with , τ ∈ (0, T₀], we get

The last term in (39)

Besides,

and

Then, taking into account (H1) – (H5), from (39) we get inequality

According to Grönwall's Lemma, we obtain

Interating (41) with respect to τ, we get the estimate

Besides, (40) and (42) for m ≥ 2 imply the estimates

and

where

Note that r²(t) = g²(t). Then, taking into account (33), we have

where M₆ is a positive constant. It follows from (42) and (38) that for m ≥ 3

By using (45) and the assumption M₄T₀ < 1, we can show the estimate

Due to (42)

Besides, (43) and (44) for m ≥ 2 imply the estimates

and

And, therefore,

and

and for k ∈ ℕ, m ≥ 2

Besides, we square both sides of equality (36), taking into account the hypotheses (H5) and obtain

From (53), we can conclude that:

Therefore,

It follows from (46), (50), (51), (52), and (55) that for any ε > 0, there exists m₀ such that for all k, m ∈ ℕ, m > m₀, the following inequalities hold:

are true. Hence, the sequence is fundamental in and in C([0, T₀]), is fundamental in and in , and is fundamental in and in . Therefore, as m → ∞

Theorem 3.2 implies the following estimate

and, by virtue of (56) , where M₇ is independent on m, and therefore the right-hand side of (57) is bounded with the constant, independent on m. Hence, we can select a subsequence of sequence (we preserve the same notation for this subsequence), such that

Taking into account (56) and (58), from (33), (34) we get that the pair of functions (u(x, t), g(t)) satisfies (27) and (26). By virtue of Lemma 4.2 (u(x, t), g(t)) is a solution of the problem (2–5) in Q_T0.

II. Uniqueness of solution of the problem (2–5), with T ≤ T₀.

Assume that (u₍₁₎(x, t), g₍₁₎(t)) and (u₍₂₎(x, t), g₍₂₎(t)) be two solutions of problem (2–5). Then the pair of functions , where ũ(x, t) = u₍₁₎(x, t)−u₍₂₎(x, t), satisfies the conditions ũ(x, 0) ≡ 0, ũ_t(x, 0) ≡ 0, the equality

for all and the equality

holds.

After choosing v = ũ_t in (59) we get

It is easy to get from (60) and (H5) inequalities

From (61) by the same way as from (39) we got (42), we find the following estimate

and taking into account (62) from (63), we obtain Since M₄T₀ < 1, we conclude that hence, u₍₁₎ = u₍₂₎ in Q_T0. Then (62) implies and, therefore, g₍₁₎(t) ≡ g₍₂₎(t) on [0, T₀].

III. Let now T > 0 be arbitrary number.

Let us divide the interval [0, T] into a finite number of intervals [0, T₁], [T₁, 2T₁], …, [(N − 1)T₁, NT₁], where NT₁ = T, and T₁ ≤ T₀. According to I and II, there exists a unique solution (u₁(x, t), g_0,1(t)) to the problem (2–5) in the domain Q_T1.

Now, we will prove that there exists a unique solution in the domain Q_{T1, 2T1}: = Ω × (T₁; 2T₁) for the problem for the Equation 2 with conditions (4) and (5) as t ∈ [T₁; 2T₁], and with the initial condition u(x, T₁) = u₁(x, T₁), u_t(x, T₁) = u_1t(x, T₁), and x ∈ Ω.

Let us change the variables t = τ+T₁, τ ∈ [0;T₁] in this problem. We will denote G₀(τ) = g(τ+T₁), U(x, τ) = u(x, τ+T₁), and For the pair (U(x, τ), G₀(τ)) we obtain the problem:

It is obvious that all coefficients of the Equation 64 and the functions E⁽¹⁾(τ) satisfy the same conditions as the functions from (2) and (5). According to I and II, there exists a unique solution to the problem (64–67) in Q_T1, and, thus for the problems for the Equation 2 with conditions (4) and (5) as t ∈ [T₁; 2T₁] and with the initial condition u(x, T₁) = u₁(x, T₁), u_t(x, T₁) = u_1t(x, T₁), and x ∈ Ω, in the domain Q_{T1, 2T1}. Denote it by (u₂(x, t), g_0,2(t)). By following similar reasoning on the intervals [2T₁; 3T₁], …, [(N − 1)T₁; NT₁], we can prove the existence and uniqueness of weak solutions (u_k(x, kt), g_0,k(t)), k = 3, …, N, in the domain Q_{(k−1)T1,kT1}: = Ω × ((k − 1)T₁, kT₁) of the inverse problem for the Equation 2 with conditions (4) and (5) as t ∈ [(k − 1)T₁; kT₁] and the initial condition u(x, (k − 1)T₁) = u_k−1(x, (k − 1)T₁), x ∈ Ω. It is clear that a pair of functions (u(x, t), g₀(t)), where

is a solution for the problem (2–5) in the domain Q_T.

IV. The uniqueness of solution is proved similar as in II, III: Assume that (u₍₁₎(x, t), g₍₁₎(t)) and (u₍₂₎(x, t), g₍₂₎(t)) be two solutions of problem (2–5). Then the pair of functions , where ũ(x, t) = u₍₁₎(x, t)−u₍₂₎(x, t), satisfies the conditions ũ(x, 0) ≡ 0, ũ_t(x, 0) ≡ 0, the equality

for all , τ ∈ (0, T], and the equality

holds.

Let us divide the interval [0, T] into a finite number of intervals [0, T₁], [T₁, 2T₁], …, [(N − 1)T₁, NT₁], where NT₁ = T, and T₁ ≤ T₀.

Let us choose τ ∈ [0, T₁] in (68). After choosing here v = ũ_t, we get

It is easy to get from (69) and (H5) inequalities

From (70), by the same way as from (39), we got (42). We find the following estimate:

and taking into account (71) from (72), we obtain Since M₄T₁ < 1, we conclude that hence, u₍₁₎ = u₍₂₎ in Q_T1. Then (71) implies and, therefore, g₍₁₎(t) ≡ g₍₂₎(t) on [0, T₁].

Let us choose τ ∈ [0, 2T₁] in (68). After choosing here v = ũ_t, we get

Note that ũ ≡ 0 in Q_T1 and on [0;T₁], therefore, from (73) it follows that

It is easy to get from (69) and (H5) inequalities

From (74), by the same way as from (39), we got (42). We find the following estimate

and taking into account (75) from (76), we obtain Since M₄T₁ < 1, we conclude that hence, u₍₁₎ = u₍₂₎ in Q_{T1, 2T1}. Then, (75) implies and, therefore, g₍₁₎(t) ≡ g₍₂₎(t) on [T₁, 2T₁]. Therefore, u₍₁₎ = u₍₂₎ in Q_2T1, g₍₁₎(t) ≡ g₍₂₎(t) on [0, 2T₁].

Considering τ ∈ [0, 3T₁], …   , τ ∈ [0, NT₁] in (68), by the same arguments, we find that u₍₁₎ = u₍₂₎ in Q_{(k−1)T1,kT1}, g₍₁₎(t) ≡ g₍₂₎(t) on [(k − 1)T₁, kT₁], k = 1, 2, …, N. Therefore, u₍₁₎ = u₍₂₎ in Q_T, g₍₁₎(t) ≡ g₍₂₎(t) on [0, T].

5 Conclusions

In this study, we have derived the necessary conditions for the existence and the uniqueness of the solution for the initial-boundary value problem, as well as the inverse problem, for semilinear hyperbolic equation of the third order with an unknown parameter in its right-hand side function. To determine the unknown function, an additional integral-type overdetermination condition have been introduced. These results were obtained by utilizing the properties of the solutions to the initial-boundary value problem and the method of successive approximations.

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